Originally Posted by

**skhan** how would I compute the answers to the following questions:

A supermarket manager finds that 20% of his customers buy brand X cornflakes. The other 80% buy brand Z.

a) Ina ramdom sample of 10, what is the probability that at least 4 customers buy brand X? [ANS:0.121]

Here we are dealing with the binomial distribution, where the probability of

$\displaystyle R$ successes in $\displaystyle N$ trials is:

$\displaystyle

P(R,N)={N \choose R}p^R (1-p)^{N-R}

$,

where $\displaystyle p$ is the probability of a success on a single trial, and:

$\displaystyle {N \choose R}=\frac{N!}{R!(N-R)!}$

is a binomial coefficient.

Now the probability $\displaystyle Prb(4+)$ that at least 4 customers out of 10 buy brand X is

1 minus the probability that 3 or less customers buy brand X. So:

$\displaystyle

Prb(4+)=1-P(0,10)-P(1,10)-P(2,10)-P(3,10)

$

$\displaystyle

=1-{10 \choose 0}0.2^00.8^10-{10 \choose 1}0.2^1 0.8^9$$\displaystyle -{10 \choose 2}0.2^2 0.8^8-{10 \choose 3}0.2^3 0.8^7

$

$\displaystyle

=1-0.1074-0.2684-0.3020-0.2013=0.1209

$

RonL